do the particles ever collide in QED

If you look up things like the Stern-Gerlach experiment, you may find the term space quantization mentioned. It simply means that an external field can create a variety of different energy states out of what were previously states of the same energy. Typically one gets this with the angular momentum of charged particles, as this creates magnetic moments which interact with electric or magnetic fields. QM tells you the angular momenta cannot adopt any orientation they like with respect to such a field, only particular orientations, associated with particular energy levels. (Having said all this I find relatively few references to the term "space quantization" on the internet. It may be that it is a term that has fallen out of favour since the 1970s when I learnt all this stuff.)

Since the spin quantum number of the electron is 1/2, its spin angular momentum, s, = h/2pi √(1/2.(1.2 + 1)), i.e. just like other forms of quantized angular momentum. However its allowed projections, m(s), along a z axis defined by an external field, are + 1/2 h/2pi or - 1/2 h/2pi.

All I'm saying is the quantum number m(s) only has significance in the presence of a field felt by the electron in question, because it is a projection of s, in a direction in space. The Pauli principle says no two electrons in an atom can have the same set of quantum numbers and of course, in an atom, m(s) is one of these, because the orbital angular momentum of the electron in an atomic orbital creates such a field. So you get different energies depending on whether the spin is up or down with respect to the orbital angular momentum, hence doublets etc.

I think I've got this right - but it was 40 years ago, I admit.

 

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Yes, I think it's important to keep in mind that the quantum numbers we associate with electrons in an atom arise from the nature of the bound state it is in (i.e. bound by a spherical Coulomb potential, due to the nucleus). These quantum numbers, n, l, m(l) and then, separately, when you take electron spin into account, m(s), are not intrinsic to the electron as a particle. They arise due to the constraints the electron experiences in an atom.

Its spin quantum number, s = 1/2, is however intrinsic to the electron, regardless of the state it is in.

If you put it in a field, you constrain it and then m(s) becomes defined.

 

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So do I, and I thank you both for your trouble

However, another question, if I may

I am having a problem with this. Is the first "state" literally a "quantum state" and he second an eigenstate, or have I misunderstood?

I fell pretty sure that here by "many states" you are referring to what exchemist would call atomic orbitals. These are almost by definition discrete eigenstates

If my last conjecture is correct, the concept of an orbital is redundant for an unbound electron, obviously

 

According to my understanding, the quantum states are all eigenstates, i.e. solutions to the time-independent form of Schroedinger's equation. They take the form they do because of the form of the Hamiltonian. The best known (to me as a chemist anyway) are in an atom where the constraining influence represented in the Hamiltonian is a spherical Coulomb potential and what results is atomic orbitals. With other forms of constraint, the Hamiltonian looks different and different sorts of eigenstates will result. (I recall at university various cases of this sort, e.g. the "particle in a box" with a square potential, the harmonic oscillator with a parabolic potential, etc.) In the case of a "free" electron there are no constraints at all and any energy is permitted to it, i.e. the "eigenstates" merge into a continuum. If I recall correctly, one actually sees this in spectroscopy, as the spectral lines converge to an eventual continuum of absorption or emission, corresponding to ionisation of the electron from the atom.

The harmonic oscillator is interesting as it gives a (simplified) insight into how the vibrations of a bond in a molecule are quantised, as seen in the IR spectrum. This is another example of a condition in which the concept of orbitals doesn't apply, yet Schroedinger's equation still does apply and there are resulting "quantum states" or "eigenstates".

 

Hmmm.

Sorry to be a boring mathematician, but I have a slight quarrel with this

In my understanding, in quantum mechanics, the state of a system is given by an element in the Hilbert space of square integrable functions called a state function, which from the preceding is a "state vector" and is commonly denoted by \(\psi\). Again, according to this theory, the so-called "observables" are given by some Hermitian operator acting on this state function/vector.

And the possible measurement values of this Hermitian operator acting on our state function/vector are given by the eigenvalues as functions of the eigenvectors = eigenstates.

In symbols \(H \psi = \lambda |\psi \rangle\) where under the action of \(H\) on each state function/vector one associates a (possible) multiplicity of eigenstates \(|\psi \rangle\) each with its own (possible) multiplicity of eigenvalues \(\lambda\)

But now I get a bit lost - the Pauli spin matrices are clearly Hermitian, and since any operator has a matrix representation, maybe (?) I can think of, say \(\sigma_z\) as James R suggested as being an operator that takes the "state" of the electron to an eigenstate/eigenvector with its associated eigenvalues And since this matrix/operator has eigenvalues +1 and -1, I need to assert that the associated eigenstate/eigenvector is \(\frac{1}{2}\hbar\).

I do not know how to do this,always assuming of course that the above is not pure gibberish

 

I recall something in a course about the eigenvalues; for a single electron you won't "see" +1 and -1, you need two electrons. Why this is I have no idea just now, but it might come back to me. I do know it has to do with measurement operators.

Ah yes, the inner product in \( \mathbb C^2 \) can only be 0 or 1 for a single particle, two particles gives you -1 as well. I think that's it.

 

Well it isn't. See this? See this bit?

"The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves."

Bump an electron out of its orbital, and guess what? It still exists as a standing wave. As for the Pauli exclusion principle, think on this: two ocean waves can ride over one another, but two whirlpools can't overlap.

 

Now you are simply embarrassing yourself. You are disagreeing with a claim by posting a sentence that is irrelevant to the claim. In your mad rush to play the role of "physics expert", you are simply spouting nonsense more directly than normal.

We can't think of whirlpools or waves because they don't apply; as people demonstrated in your question dodging thread, thinking of these kinds of fluid dynamics produces gross inconsistencies with quantum mechanics. Even though you have yet to produce a model of quantum particles that one can use to do physics, what you have established is that using fluid dynamics as you suggest has a fatal flaw.

 

It may that as a chemist I'm taking an atom-based view of this, but I did think all stable states of a quantum system had to be eigenstates (=defined by eigenfunctions) of the time-independent Schroedinger equation for the system. Does what you are saying conflict with this? But I will be the first to admit I am not nimble with the maths of this any longer.

I can't help much with the spin question. We didn't really get much into the spin operators at university and tended to treat spin as a bit of a bolt-on extra for its effects on atomic and molecular structure and properties. Looking at James' post it may be that clears it up.

 

No they didn't, and I didn't dodge questions. The photon is a boson, it has an E=hc/λ wave nature, and when you google vortex fermion there's a lot of hits. The Pauli exclusion principle doesn't work through magic you know. And putting it down to "quantum numbers" doesn't get to the bottom of it. But when you keep reading and see total angular momenta numbers, you start to get the drift.

 

Farsight, you are making this up. An unbound electron is not quantised, so it makes no sense to speak of standing waves in such a context.

You will know from experience that a standing wave arises when you have a trapped wave, reflecting from the extremities of some sort of confinement. To use the sort of language you seem to deal in, " No confinement, no standing wave".

 

Look, you may be too mentally damaged to even understand what an answer is anymore, or you may just be lying.

All you ever do is dodge questions. You at least admited that you have no mathematics to support your claims. That on its own means that you cannot have an explanation. However, you keep going with pictures that even you say can't provide an exact description.

Sometimes, you clearly lie and say that all you are doing is presenting the wave theory of matter. Then other times you commit to truly crazy things like all electromagnetism is curved space and that curved space makes a photon travel in a vortex. Then you go back to lying and claiming that denying your curved space idea is denying the wave nature of matter.

You have proven that you can google and that you can cut and paste. But you have also shown that you will lie when it suits your purpose and that you do not understand much about physics. This is the hard evidence that you have given us.

What does this kind of behavior say about someone who is supposedly trying to fix science education because of his children? If they had children, what kind of math-free, Einstein worshipping future would you give them?

 

Farsight - please don't interrupt when the grown-ups are talking

But look....

I now realize where my confusion lay - the above is nonsense. What we have here are two operators - the spin angular momentum and the spin matrix operators, each with their own eigenvalues.

So for example if I take \(m_z\) as the spin angular momentum in the (x,y) plane, and \(\sigma_z\) as one of the Pauli matrix operator, I can show, with some difficulty, the eigenvalues for \(m_z\) are \(\pm\frac{1}{2} \hbar\) and those for \(\sigma_z\) are \( \pm1\).

The calculation I used made use of the commutators of each (I looked them up!!) and the creation/annihilation operators of P.A.M Dirac (I looked these up too). These calculations - at least in my hands - are long and rather inelegant.

But anyway, this means that in general I may have the operator equality \(\mathbf{m}= \frac{1}{2}\hbar \mathbf{\sigma}\)

 

You don't know what you're talking about.

I'm absolutely not making it up. Light is quantised, the photon has an E=hf quantum nature, and we make an electron (and a positron) in gamma-gamma pair production which converts two field-variations into two standing fields. Standing wave, standing field. They have the opposite chirality and the opposite angular momentum, which is conserved. Electron-positron annihilation reverses the process.

There is a confinement. The electron is akin to a photon "in a box of its own making". That's why we can diffract electrons. The wave nature of matter is a certainty. So if you've got an electron sitting there in front of you, you know it's a wave with electromagnetic characteristics, you know it isn't departing thataway → at c, so you know that it's somehow trapped. You also know about the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". So you can be certain that there's something going round. What isn't a certainty is the precise way it works: there is no electron model in the Standard Model. But even without that you should see that this is why the inertia of a body is a measure of its energy-content. And that annihilation is like opening one box with another, whereupon you have two radiating bodies losing mass. Only they lose all of it, then they're not there any more. It's conceptually very simple. Don't reject it because it wasn't in your textbook. Because it will be.

 

Only according to your pseudo-hypothesis. Since you have no equations for your idea, and hence no experimental support, we must simply ignore your claims in this regard.

 

Golly. This is getting almost medical. What is the point in making up all this stuff, when there is already a clear explanation, underpinned by almost a century of mathematics and experiment, that works?

 

Farsight, these are the same sort of arguments that resulted in another thread in Alternative Theories. Is it your intent to try and drive every thread you can in that direction?

 

OK now I'm intrigued - and my textbooks don't do Pauli spin operators. The physical significance of m(z) is well-known to us chemists of course. Do the eigenvalues of σ(z) have physical significance?